Integrating gravity(g)= -9.8m/s

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Integrating gravity at -9.8 m/s² leads to equations of motion under constant acceleration. The discussion includes attempts to set integral limits and derive motion equations, with specific focus on velocity and displacement. Participants clarify the distinction between Newton's physics and mathematics, emphasizing force as mass times acceleration. There are requests for corrections on derived equations and a need for assistance with LaTeX formatting. The conversation highlights challenges in understanding and applying these concepts accurately.
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correct me if I am wrong...

By integrating gravity(g)= -9.8m/s...you get the motion equations with constant acceleration

I didnt know how to set the limits of the integral...LATEX IS TUFF
d\vec{a}=(-g)dt 1

\int_{\vec{V}_0}^{\vec{V}}dv=\int_{t_0}^{t_1}{-g}dt 2

\Delta{\vec{v}}=(-g)t 3

\vec{v}dt=d\vec{r} 4

\d\vec{r}=(\vec{v}_0+(-g)t)dt 5

\intd\vec{r}=\int(\vec{v}_0+(-g)tdt 6

\Delta\vec{r}=\vec{v}_0t+1/2(-g)t^2 7

\Delta{\vec{v}}=\int{\vec{v}_0}dt 8


HOLD ON A SEC IM TRYN TO LATEX...Can anyone prove the force equations?

HI merons dad
 
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Are you talking about Newton the physicist or Newton the mathematician?

Physics: force = mass times acceleration or change in momentum per unit time

acceleration = time rate of change of velocity

Mathematics: a = \frac {dv}{dt}
 
Gravity

Did I make some mistakes...?..i think i fixed them all. Can anyone show how to derive Torqu?

\Delta{\vec{v}}=(-g)t 3

\vec{v}dt=d\vec{r} 4

d\vec{r}=(\vec{v}_0+(-g)t)dt 5

\intd\vec{r}=\int(\vec{v}_0+(-g)tdt 6

\Delta\vec{r}=\vec{v}_0t+1/2(-g)t^2 7

\Delta{\vec{v}}=\int{\vec{v}_0}dt 8
 
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Oooops I ment to put that on the physics posts not calculus :frown: :frown:
 
how did he figure it out..experiment?
 
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No English!
 
Alem, there appears to be a problem with your line #8.
 
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